Chapter 15 Energy and Chemical Change Chemical reactions usually absorb or release energy. Section 1: Energy Section 2: Heat Section 3: Thermochemical Equations Section 4: Calculating Enthalpy Change Section 5: Reaction Spontaneity Section 1: Energy can change form and flow, but it is always conserved Section 2: The enthalpy change for a reaction is the enthalpy of the products minus the enthalpy of the reactants. Section 3: Thermochemical equations express the amount of heat released or absorbed by chemical reactions. Section 4: The enthalpy change for a reaction can be calculated using Hess s law. Section 5: Changes in enthalpy and entropy determine whether a process is spontaneous. What is energy? How do potential and kinetic energy differ? How can chemical potential energy be related to the heat lost or gained in chemical reactions? How is a calorimeter used to measure energy that is absorbed or released? What do enthalpy and enthalpy change mean in terms of chemical reactions and processes? How are thermochemical equations for chemical reactions and processes written? How is energy lost or gained during changes of state? How is the heat that is absorbed or released in a chemical reaction calculated? 1
continued How is Hess s law applied to calculate the enthalpy change in a reaction? What is the basis for the table of standard enthalpies of formation? What is the enthalpy change for a reaction using standard enthalpies of formation data? What is the difference between spontaneous and nonspontaneous processes? How do changes in entropy and free energy determine the spontaneity of chemical reactions and other processes? temperature pressure combustion reaction allotrope vaporization New energy law of conservation of energy chemical potential energy heat New continued calorie joule specific heat calorimeter thermochemistry system surroundings universe enthalpy enthalpy (heat) of reaction New continued thermochemical equation enthalpy (heat) of combustion molar enthalpy (heat) of vaporization molar enthalpy(heat) of fusion Hess s law New continued standard enthalpy (heat) of formation spontaneous process entropy second law of thermodynamics free energy Section 1: Energy Energy can change form and flow, but it is always conserved 2
What is energy? How do potential and kinetic energy differ? How can chemical potential energy be related to the heat lost or gained in chemical reactions? temperature New energy law of conservation of energy chemical potential energy heat calorie joule specific heat The Nature of Energy Energy is the abiltiy to do work or produce heat. Energy exists in two basic forms: potential energy and kinetic energy. Potential energy is energy due to composition or position. Kinetic energy is energy of motion. The Nature of Energy The law of conservation of energy states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed. Also known as the first law of thermodynamics. Chemical potential energy is energy stored in a substance because of its composition. Chemical potential energy is important in chemical reactions. Heat is energy that is in the process of flowing from a warmer object to a cooler object. q is used to symbolize heat. 3
Measuring Heat A calorie is defined as the amount of energy required to raise the temperature of one gram of water one degree Celsius. The energy content of food is measured in Calories, or 1000 calories. Also known as a kilocalorie. A joule is the SI unit of heat and energy, equivalent to 0.2390 calories. Relationships among Energy Units Relationship Conversion Factor 1 J =0.2390 cal 1 J/0.2390 cal OR 0.2390 cal/1 J 1 cal=4.184 J 1 cal/4.184 J OR 4.184 J/1 cal 1 Cal = 1 kcal 1 Cal/1000 cal OR 1000 cal/1 Cal 1 Cal = 4.184 kj 1 Cal/4.184 kj OR 4.184 kj/1 Cal CONVERT ENERGY UNITS Use with Example Problem 1. Problem A breakfast of cereal, orange juice, and milk might contain 230 Calories. Convert this energy to joules. Response ANALYZE THE PROBLEM You are given an amount of energy in nutritional Calories. You must convert Calories to calories, and then convert calories to joules. KNOWN Amount of enegy = UNKNOWN Amount of energy = 230 Calories? J Convert Calories to calories Apply the relationship 1 Calorie = 1000 cal 1000 230 =2.3 10 1 Convert calories to joules Apply the relationship 1 calorie = 4.184 joules 2.3 10 4.184 =. 1 EVALUATE THE ANSWER The minimum number of significant figures in the conversion is two, and the answer correctly has two digits. A value on the order of 10 5 or 10 6 is expected because the given number of kilocalories is of the order of 10 2 and it must be multiplied by 10 3 to convert it to calories. Then, the calories must be multiplies by a factor of 4. Therefore, the answer is reasonable. Specific Heat The specific heat of any substance is the amount of heat required to raise one gram of that substance one degree Celcius. Some objects required more heat than others to raise their temperature. Specific Heats at 298 K (25⁰C) Substance c in J/gK Aluminum 0.897 Bismuth 0.123 Copper 0.386 Brass 0.380 Gold 0.126 Lead 0.128 Silver 0.233 Tungsten 0.134 Zinc 0.387 Mercury 0.140 Alcohol(ethyl) 2.44 Water 4.186 Ice (-10 C) 2.05 Granite 0.803 Concrete 0.840 Specific Heat Calculating heat absorbed and released Equation for calculating heat q=mxc xδt q represents the heat absorbed or released m represents the mass of the sample in grams c represents the specific heat of the substance ΔT T is the change in temperature in Celsius or Kelvin The quantity of heat absorbed or released by a substance is equal to the product of its specific heat, the mass of the substance, and the change in its temperature. 4
CALCULATE SPECIFIC HEAT Use with Example Problem 2. Problem In the construction of bridges and skyscrapers, gaps must be left between adjoining steel beams to allow for the expansion and contraction of the metal due to heating and cooling. The temperature of a sample of iron with a mass of 10.0 g changed from 50.4 ⁰C to 25.0 ⁰C with the release of 114 J. What is the specific heat of iron? Response ANALYZE THE PROBLEM You are give the mass of the sample, the initial and final temperatures, and the quantity of heat released. You can calculate the specific heat of iron by rearranging the equation that relates these variables to solve for c. KNOWN Energy released = 114 J Mass of iron = 10.0 g Fe UNKNOWN Specific heat of iron, T i = 50.4 ⁰C T f = 25.0 ⁰C c =? J/g⁰C Calculate ΔT. ΔT=50.4 25.0 =25.4 State the equation for calculating heat = Δ solve for c = Δ Substitute q=114 J, m=10.0 g and T = 25.4⁰C 114 = 10.0 25.4 solve for c =. What is energy? How do potential and kinetic energy differ? How can chemical potential energy be related to the heat lost or gained in chemical reactions? energy law of conservation of energy chemical potential energy heat calorie joule specific heat Section 2: Heat The enthalpy change for a reaction is the enthalpy of the products minus the enthalpy of the reactants. How is a calorimeter used to measure energy that is absorbed or released? What do enthalpy and enthalpy change mean in terms of chemical reactions and processes? 5
pressure New calorimeter thermochemistry system surroundings universe enthalpy enthalpy (heat) of reaction Calorimetry A calorimeter is an insulated device used for measuring the amount of heat absorbed or released in a chemical reaction or physical process. CALCULATE SPECIFIC HEAT Use with Example Problem 3. Problem A piece of metal with a mass of 4.68 g absorbs 256 J of heat when its temperature increases by 182 ⁰C. What is the specific heat of the metal? Response ANALYZE THE PROBLEM You are give the mass of the sample, the change in temperature, and the quantity of heat absorbed. You can calculate the specific heat of the metal by rearranging the equation that relates these variables to solve for c. KNOWN Energy absorbed = 256 J Mass of metal = 4.68 g UNKNOWN Specific heat T = 182 ⁰C c =? J/g⁰C State the equation for calculating heat = Δ solve for c = Δ Substitute q=256 J, m=4.68 g and T = 182⁰C 256 = 4.68 182 =. EVALUATE THE ANSWER The three quantities used in the calculation have three significant figures, and the answer is correctly stated with three digits. The calculations are correct and yield the expected unit. Chemical Energy and the Universe Thermochemistry is the study of heat changes that accompany chemical reactions and phase changes. The system is the specific part of the universe that contains the reaction or process you wish to study. The surroundings are everything else other than the system in the universe The universe is defined as the system plus the surroundings 6
Chemical Energy and the Universe Chemists are interested in changes in energy during reactions. Enthalpy is the heat content of a system at constant pressure. Enthalpy changes for exothermic reactions are always negative. Enthalpy changes for endothermic reactions are always positive. Chemical Energy and the Universe Enthalpy (heat) of reaction is the change in enthalpy during a reaction. Symbolized as H rxn. H rxn = H final H initial H rxn = H products H reactants Chemical Energy and the Universe How is a calorimeter used to measure energy that is absorbed or released? What do enthalpy and enthalpy change mean in terms of chemical reactions and processes? calorimeter thermochemistry system surroundings universe enthalpy enthalpy (heat) of reaction 7
Section 3: Thermochemical Equations Thermochemical equations express the amount of heat released or absorbed by chemical reactions. How are thermochemical equations for chemical reactions and processes written? How is energy lost or gained during changes of state? How is the heat that is absorbed or released in a chemical reaction calculated? combustion reaction New thermochemical equation enthalpy (heat) of combustion molar enthalpy (heat) of vaporization molar enthalpy(heat) of fusion Writing Thermochemical Equations A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products, and energy change The enthalpy (heat) of combustion of a substance is the enthalpy change for the complete burning of one mole of the substance. Standard Enthalpies of Combustion Substance Formula ⁰ΔH comb (kj/mol) Sucrose (table sugar) C 12 H 22 O 11 (s) -5644 Octane (in gasoline) C 8 H 18 (l) -5471 Glucose (simple sugar) C 6 H 12 O 6 (s) -2808 Propane C 3 H 8 (g) -2219 Methane (found in natural gas) CH 4 (g) -891 8
Changes of State Changes of State Molar enthalpy (heat) of vaporization refers to the heat required to vaporize (or boil) one mole of a liquid substance Molar enthalpy (heat) of fusion refers to the heat required to melt one mole of a solid substance Standard Enthalpies of Vaporization and Fusion Substance Formula ⁰ΔH vap (kj/mol) ⁰ΔH fus (kj/mol) Water H 2 O 40.7 6.01 Ethanol C 2 H 5 OH 38.6 4.94 Methanol CH 3 OH 35.2 3.22 Acetic Acid CH 3 COOH 23.4 11.7 Ammonia NH 3 23.3 5.66 CALCULATING ENTHALPY CALCULATING ENTHALPY Use with Example Problem 4. Problem A bomb calorimeter is useful for measuring the energy released in combustion reactions. The reaction is carried out in a constant-volume bomb with a high pressure of oxygen. How much heat is released with 54.0 g of glucose (C 6 H 12 O 6 ) is burned according to this equation? Response ANALYZE THE PROBLEM You are given a mass of glucose, the equation for the combustion of glucose, and H comb. You must convert grams of glucose to moles of glucose. Because the molar mass of glucose is more than three times the mass of the glucose burned, you can predict that the energy released will be less than one third of H comb. Multiply moles of C 6 H 12 O 6 by the enthalpy of combustion, H comb. Multiply moles of glucose by. 0.300 2808 = 1 EVALUATE THE ANSWER All values in the calculation have three significant figures, so the answer is correctly stated with three digits. As predicted, the released energy is less than one third of H comb. C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 O H comb = -2808 kj KNOWN UNKNOWN Mass of glucose = q =? kj 54.0 g C 6 H 12 O 6 H comb = -2808 kj Convert grams of C 6 H 12 O 6 to moles of C 6 H 12 O 6 Divide by the molar mass. 1 54.0 180.18 =0.300 9
Combustion Reactions Combustion is the reaction of a fuel with oxygen. Food is the fuel in combustion reactions in biological systems. How are thermochemical equations for chemical reactions and processes written? How is energy lost or gained during changes of state? How is the heat that is absorbed or released in a chemical reaction calculated? thermochemical equation enthalpy (heat) of combustion molar enthalpy (heat) of vaporization molar enthalpy(heat) of fusion Section 4: Calculating Enthalpy Change The enthalpy change for a reaction can be calculated using Hess s law. How is Hess s law applied to calculate the enthalpy change in a reaction? What is the basis for the table of standard enthalpies of formation? What is the enthalpy change for a reaction using standard enthalpies of formation data? 10
allotrope New Hess s law standard enthalpy (heat) of formation Hess s Law Hess s law states that if you can add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction. CALCULATE SPECIFIC HEAT KNOWN a 2H 2 + O 2 2 H 2 O H = -572 kj b H 2 + O 2 H 2 O 2 H = -188 kj CALCULATE SPECIFIC HEAT Use with Example Problem 3. Problem Use thermochemical equations a and b below to determine H for the decomposition of hydrogen peroxide (H 2 O 2 ), a compound that has many uses from bleaching hair to powering rocket engines. 2H 2 O 2 2H 2 O + O 2 a 2H 2 + O 2 2 H 2 O H = -572 kj b H 2 + O 2 H 2 O 2 H = -188 kj UNKNOWN H =? kj H 2 O 2 is a reactant Reverse equation b and change the sign of H H 2 O H 2 2 + O 2 H = + 188 kj Two moles of H 2 O 2 are needed Multiply Reversed equation b and H by 2 2 H 2 O 2 2H 2 + 2O 2 H = + 376 kj Add equations a and new b, canceling any terms that appear on opposite sides Write equation a 2H 2 + O 2 2 H 2 O H = 572 kj Write new equation b 2 H 2 O 2 2H 2 + 2O 2 H = + 376 kj Add the equations and cancel. 2 H 2 O 2 2H 2 + 2O 2 H = + 376 kj 2H 2 + O 2 2 H 2 O H = 572 kj EVALUATE THE ANSWER The two equations produce the desired equation. All values are accurate to the ones place, so H is correctly stated. Response ANALYZE THE PROBLEM You have been given two chemical equations and their enthalpy changes. These two equations contain all the substances found in the desired equation. Add the H. 2H 2 O 2 2H 2 O + O 2 H = 196 kj 11
Standard Enthalpy (Heat) of Formation The standard enthalpy (heat) of formation is defined as the change in enthalpy that accompanies the formation of one mole of the compound in its standard state from its elements in their standard states. Elements in their standard states have a H f ⁰ of 0.0 kj. The formation of compounds are placed above or below elements in their standard states. Standard Enthalpy (Heat) of Formation Standard enthalpies of formation can be used to calculate the enthalpies for many reactions under standard conditions by using Hess s law. The summation equation: H rxn ⁰ = Σ H f ⁰(products) -Σ H f ⁰(reactants) Select Standard Enthalpies of Formation Substance Formation Equation ⁰ΔH f ⁰(kJ/mol) H 2 S (g) H 2 (g) + S(s) H 2 S(g) -21 HF(g) 0.5H 2 (g) + 0.5F 2 (g) HF(g) -273 SO 3 (g) S(s) + 1.5O 2 SO 3 (g) -396 SF 6 (g) S(s) + 3F 2 SF 6 (g) -1220 ENTHALPY CHANGES Use with Example Problem 6. Problem Use Standard Enthalpies of Formation to calculate H rxn ⁰ for the combustion of methane. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) Response ANALYZE THE PROBLEM You are given an equation and asked to calculate the change in enthalpy. The formula H rxn ⁰ = Σ H f ⁰(products) -Σ H f ⁰(reactants) can be used with data from table R-11, which is on the next slide. 12
ENTHALPY CHANGES Use with Example Problem 6. Problem Use Standard Enthalpies of Formation to calculate H rxn ⁰ for the combustion of methane. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) Response ANALYZE THE PROBLEM You are given an equation and asked to KNOWN H f ⁰ CO 2 = -394 kj H f ⁰ H 2 O(l) = -286 kj H f ⁰ CH 4 = -75 kj H f ⁰ O 2 = 0.0 kj UNKNOWN H rxn ⁰ =? kj Use the formula H rxn ⁰ = Σ H f ⁰(products) - Σ H f ⁰(reactants). Expand the formula to include a term for each reactant and product. Multiply each term by the coefficient of the substance in the balanced chemical equation. calculate the change in enthalpy. The formula H rxn ⁰ = Σ H f ⁰(products) -Σ H f ⁰(reactants) can be used with data from table R-11, which is on the next slide. Substitute CO 2 and H 2 O for the products, CH 4 and O 2 for the reactants. Multiply H 2 O and O 2 by 2. H rxn ⁰=( H f ⁰ CO 2 + (2 x H f⁰ H 2 O)) -( H f ⁰ CH 4 + (2 x H f⁰ O 2 )) ENTHALPY CHANGES Substitute Hf⁰ CO 2 = -394 kj, Hf⁰ H 2 O(l) = -286 kj, Hf⁰ CH 4 = -75 kj, and Hf⁰ O 2 = 0.0 kj into the equation H rxn ⁰=(-394 kj + (2 x -286 kj)) (-75 kj + (2 x 0.0 kj)) H rxn ⁰=(-966 kj) (-75 kj) = -966 kj + 75 kj = -891 kj The combustion of 1 mole CH 4 releases 891 kj. EVALUATE THE ANSWER All values are accurate to the ones place. Therefore, the answer is correct as stated. The calculated value is the same as the Enthalpy of Combustion value given on the next slide. ENTHALPY CHANGES Standard Enthalpies of Combustion Substance Formula ⁰ΔH comb (kj/mol) Sucrose (table sugar) C 12 H 22 O 11 (s) -5644 Octane (in gasoline) C 8 H 18 (l) -5471 Glucose (simple sugar) C 6 H 12 O 6 (s) -2808 Propane C 3 H 8 (g) -2219 Methane (found in natural gas) CH 4 (g) -891 How is Hess s law applied to calculate the enthalpy change in a reaction? What is the basis for the table of standard enthalpies of formation? What is the enthalpy change for a reaction using standard enthalpies of formation data? Hess s law standard enthalpy (heat) of formation 13
Section 5: Reaction Spontaneity Changes in enthalpy and entropy determine whether a process is spontaneous. What is the difference between spontaneous and nonspontaneous processes? How do changes in entropy and free energy determine the spontaneity of chemical reactions and other processes? vaporization New spontaneous process entropy second law of thermodynamics free energy Spontaneous Process A spontaneous process is a physical or chemical change that once begun, occurs with no outside intervention. Many spontaneous processes requires some energy from the surroundings to start the process. 14
Spontaneous Process Entropy is a measure of the possible ways that the energy of a system can be distributed, and this is related to the freedom of the system s particles to move and the number of ways they can be arranged. Spontaneous Process The second law of thermodynamics states that spontaneous processes always proceed in such a way that the entropy of the universe increases. Entropy is sometimes considered a measure of disorder or randomness of the particles in a system The more spread out the particles are, the more disorder Entropy changes associated with changes in state can be predicted. Entropy increases as a substance changes from a solid to a liquid to a gas. Dissolving a gas in a solvent always results in a decrease in entropy. Spontaneous Process Assuming no change in physical state occurs, the entropy of a system usually increases when the number of gaseous product particles is greater than the number of gaseous reactant particles. With some exceptions, entropy increases when a solid or liquid is dissolved in a solvent The random motion of particles of a substance increases as its temperature increases. Entropy, the Universe, and Free Energy In nature, the change in entropy tends to be positive when: The reaction or process is exothermic, which raises the temperature of the surroundings The entropy of the system increases. 15
Entropy, the Universe, and Free Energy Free energy is energy that is available to do work. If the sign of the free energy change, G⁰, is positive, the reaction is non spontaneous. If the sign of the free energy change is negative, the reaction is spontaneous. Equation for calculating heat G system = H system - T S system G system represents the free energy change H system represents the change in enthalpy T T is the temperature in Kelvin S system represents the change in entropy system The free energy released or absorbed in a chemical reaction is equal to the difference between the enthalpy change and the product of the change in entropy and the temperature. Entropy, the Universe, and Free Energy ΔH ΔS Spontaneous? Positive Negative NEVER Negative Positive ALWAYS Negative Negative Only at low temperatures Postive Positive Only at high temperatures DETERMINE REACTION SPONTANEITY Use with Example Problem 7. Problem For a process, H system = 145 kj and S system = 322 J/K. Is the process spontaneous at 382 K? Response ANALYZE THE PROBLEM You must calculate G system to determine spontaneity. KNOWN UNKNOWN T = 382 K Sign of G system =? H system = 145 kj S system = 322 J/K Convert S system to kj/k 322 1 =0.322 / 1000 Solve the free energy equation State the Gibbs Free Energy Equation G system = H system - T S system Substitute in T = 382 K, H system = 145 kj, and S system =0.322 kj/k ΔG system = 145 kj (382 K x 0.322 kj/k) Multiply and subtract G system = 145 kj 123 kj G system = 22 kj EVALUATE THE ANSWER Because G system is positive, the reaction is nonspontaneous at that temperature What is the difference between spontaneous and nonspontaneous processes? How do changes in entropy and free energy determine the spontaneity of chemical reactions and other processes? spontaneous process entropy second law of thermodynamics free energy 16